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ICSE Class 10 Physics Question 5 of 27

Solved 2024 Question Paper ICSE Class 10 Physics — Question 20

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Question 2(v)

Sumit does 600 J of work in 10 min and Amit does 300 J of work in 20 min. Calculate the ratio of the powers delivered by them.

Answer

Given,

Work done by Sumit (WS) = 600 J

Time taken by Sumit (tS) = 10 min = 10 x 60 = 600 seconds

Work done by Amit (WA) = 300 J

Time taken by Amit (tA) = 20 min = 20 x 60 = 1200 seconds

Then,

Power delivered by Sumit (PS) = Work done by Sumit (WS)Time taken by Sumit (tS)=600600=1 W\dfrac{\text {Work done by Sumit}\ (\text W_\text S)}{\text {Time taken by Sumit}\ (\text t_\text S)} =\dfrac{600}{600} = 1\ \text W

and

Power delivered by Amit (PA) = Work done by Amit (WA)Time taken by Amit (tA)=3001200=0.25 W\dfrac{\text {Work done by Amit}\ (\text W_\text A)}{\text {Time taken by Amit}\ (\text t_\text A)} =\dfrac{300}{1200} = 0.25\ \text W

So,

PSPA=10.25=4\dfrac{\text P_\text S}{\text P_\text A}=\dfrac{1}{0.25} =4